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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Solution</dfn> The general solution to the corresponding homogeneous equation</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
y^{\prime \prime}+y=0
\end{equation*}
</div>
<p class="continuation">is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
y=C_1 \cos x+C_2 \sin x.
\end{equation*}
</div>
<p class="continuation">Let</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation}
Y=u_1(x) \cos x+u_2(x) \sin x.\tag{3.7.9}
\end{equation}
</div>
<p class="continuation">Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
Y^{\prime}=-u_1 \sin x+u_2 \cos x+\underline{u_1 ^{\prime} \cos x+u_2^{\prime} \sin x}.
\end{equation*}
</div>
<p class="continuation">Choose that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation}
u_1 ^{\prime} \cos x+u_2^{\prime} \sin x=0.\tag{3.7.10}
\end{equation}
</div>
<p class="continuation">Further, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation}
Y^{\prime \prime}=-u_1^{\prime} \sin x-u_1 \cos x+u_2^{\prime} \cos x-u_2 \sin x.\tag{3.7.11}
\end{equation}
</div>
<p class="continuation">Substituting (<a href="" class="xref" data-knowl="./knowl/eq3_31.html" title="Equation 3.7.9">(3.7.9)</a>) and (<a href="" class="xref" data-knowl="./knowl/eq3_32.html" title="Equation 3.7.11">(3.7.11)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq3_33.html" title="Equation 3.7.8">(3.7.8)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
\begin{aligned}
-u_1^{\prime} \sin x-u_1 \cos x+u_2^{\prime} \cos x-u_2 \sin x+u_1 \cos x+u_2 \sin x=\tan x,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">which implies</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation}
-u_1^{\prime} \sin x+u_2^{\prime} \cos x=\tan x.\tag{3.7.12}
\end{equation}
</div>
<p class="continuation">Solve <span class="process-math">\(u_1^{\prime}\)</span> and <span class="process-math">\(u_2^{\prime}\)</span> from (<a href="" class="xref" data-knowl="./knowl/eq3_34.html" title="Equation 3.7.10">(3.7.10)</a>) and (<a href="" class="xref" data-knowl="./knowl/eq3_33_1.html" title="Equation 3.7.12">(3.7.12)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
u_1^{\prime}=\frac{(\cos x)^2-1}{\cos x},\quad u_2^{\prime}=\sin x.
\end{equation*}
</div>
<p class="continuation">Then one solution to <span class="process-math">\(u_1\)</span> and <span class="process-math">\(u_2\)</span> are respectively</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
u_1=\sin x+\ln \sqrt {\frac{1-\sin x}{1+\sin x}},\quad u_2=-\cos x.
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
Y=\left(\sin x+\ln \sqrt {\frac{1-\sin x}{1+\sin x}}\right) \cos x-\cos x \sin x=\cos x \cdot \ln \sqrt {\frac{1-\sin x}{1+\sin x}}.
\end{equation*}
</div>
<p class="continuation">Then the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_31.html ./knowl/eq3_32.html ./knowl/eq3_33.html ./knowl/eq3_34.html ./knowl/eq3_33_1.html">
\begin{equation*}
y=C_1 \cos x+C_2 \sin x+\cos x \cdot \ln \sqrt {\frac{1-\sin x}{1+\sin x}}.
\end{equation*}
</div>
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